Balls in ultrametric spaces

Question

Consider a ball $B$ in the ultrametric space $X=K^n$ where $K$ is a $p$-adic field. (Recall that a ball in $X$ is a set of the form $\lbrace x \in K^n \mid \|x\|:= \max_{i=1}^n (\mid x_1 \mid, \dots, \mid x_n \mid ) \leq \varepsilon \rbrace$ if it is centered at the origin). Consider also another smaller ball $B' \subset B$. Is it true that $B$ can be covered by $q^{nk}$ translates of $B'$ where $q$ is the cardinality of the residue field of $K$ and $k$ is a natural number?

Answer 1

By a suitable affine transformation we can be reduced to the case when $B$ is the unit ball $R^n$ where $R$ is the valuation ring of $K$. Again by translating $(0,...,0)$ to a point of $B'$ (which leave the unit ball globally invariant), we can suppose that $B'=(\pi^k R)^n$ for some $k\ge 0$. Using translations of $B'$ by a system of representaives of the quotient group $B/B'$, we see that $B$ is covered by $|B/B'|$ copies of $B'$. As $B/B'=(R/\pi^kR)^n$, it is enough to compute $|R/\pi^kR|$.

The kernel of the canonical map $R/\pi^{k}R \to R/\pi^{k-1}R$ is $\pi^{k-1}R/\pi^k R$ which is isomorphic to $R/\pi R$ the residue field of $R$. So by induction on $k$ we see that $|R/\pi^k R|=q^k$. So $|B/B'|=q^{kn}$.

Answer 2

$B^\prime$ can be empty right? Then the answer is no if $B$ is not empty.

If $B^\prime$ is not empty, then strange things could happen. I think $B^\prime$ can just contain one single point because the space is totally disconnected. So if $B$ contains two points, the answer to your question is no.